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# α^n+β^n+γ"n standard results Watch

1. Does anyone have a link/ know a formulas for roots of polynomial standard results Ie α^2+β^2+γ^2=(α+β+γ)^2-2(αβ+βγ+γα)
And also if any one knows a formula for sigma αβ^n that would be appreciated aswell
2. I don't really get what you want here but if you're looking for a formula calculating the roots of a polynomial, there's Vieta's formulas for sums and products of roots.
3. expanding (a+b+y)^2 we get (a+b+y)^2=a^2+b^2+y^2+2(ab+by+ya )

hence (a+b+y)^2-2(ab+by+ya)=a^2+b^2+y^2

There's no formula for ab^n but for cases of trinomials it's fairly easy.

suppose x^3+kx^2+ux+p=0 and you need to find a^3+b^3+y^3

ok suppose we put a into the equation (where a,b,y are roots)

we get a^3+ka^2+ua+p, do this for cases b and y

we get b^3+kb^2+ub+p and y^3+ky^2+uy+p=0

adding all these cases together we get (a^3+b^3+y^3)+k(a^2+b^2+y^2)+u(a +b+y)+3p=0.

This sorta method applies for any situation you want.

Hoped this helped!
4. (Original post by Argylesocksrox)
expanding (a+b+y)^2 we get (a+b+y)^2=a^2+b^2+y^2+2(ab+by+ya )

hence (a+b+y)^2-2(ab+by+ya)=a^2+b^2+y^2

There's no formula for ab^n but for cases of trinomials it's fairly easy.

suppose x^3+kx^2+ux+p=0 and you need to find a^3+b^3+y^3

ok suppose we put a into the equation (where a,b,y are roots)

we get a^3+ka^2+ua+p, do this for cases b and y

we get b^3+kb^2+ub+p and y^3+ky^2+uy+p=0

adding all these cases together we get (a^3+b^3+y^3)+k(a^2+b^2+y^2)+u(a +b+y)+3p=0.

This sorta method applies for any situation you want.

Hoped this helped!
Thanks it did!

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